# What is the probability of getting STRAIGHT FLUSH in a $13$-card poker game?

What is the probability of getting STRAIGHT FLUSH in a $13$-card poker game?

Here is my attempt:

A straight flush is five cards in sequence and of the same suit, but not ace king queen jack ten.

The required probability is $$\dfrac{9 \cdot {4 \choose 1}}{{52 \choose 13}} = \dfrac{36}{635013559600} = \dfrac{9}{158753389900} \approx 5.66917028081678777\ldots \cdot{10}^{-11}.$$

My questions are:

(1) Is this probability computation correct?

(2) If my computation is not correct, where is/are the error(s) and what hint can you give towards rectifying that error(s)?

• As in the question about $4$ of a kind, the count is very much too small. It does not take into account the other $8$ cards. Also, do two straight flush count? An Inclusion/Exclusion argument will work. Apr 16 '16 at 1:55
• @AndréNicolas, yes - two straight flushes will count. Will an Inclusion/Exclusion argument similar to that in the question about $4$ of a kind, work? Apr 16 '16 at 2:16
• There are $13 \choose 5$ ways to obtain a given straight flush among the 13 card hand, so multiply your result by that Apr 16 '16 at 2:20
• btw, who plays 13-card poker? Apr 16 '16 at 2:23
• @frogfanitw, 13-card poker is also known as Chinese poker. Apr 16 '16 at 2:24

$$\frac {\binom{9}{1} \binom{4}{1} \binom{46}{8} } { \binom{52}{13} } = 0.01479 \approx 1 / 67.6$$
$$\frac {\binom{9}{1} \binom{4}{2} \binom{40}{3} } { \binom{52}{13} } = 0.00000084 \approx 1 / 1190234$$